C Program for Longest Palindromic Subsequence | DP-12

Given a sequence, find the length of the longest palindromic subsequence in it.

As another example, if the given sequence is “BBABCBCAB”, then the output should be 7 as “BABCBAB” is the longest palindromic subsequence in it. “BBBBB” and “BBCBB” are also palindromic subsequences of the given sequence, but not the longest ones. 

1) Optimal Substructure: 
Let X[0..n-1] be the input sequence of length n and L(0, n-1) be the length of the longest palindromic subsequence of X[0..n-1]. 
If last and first characters of X are same, then L(0, n-1) = L(1, n-2) + 2. 
Else L(0, n-1) = MAX (L(1, n-1), L(0, n-2)). 

Following is a general recursive solution with all cases handled.  

C

// C program of above approach
#include <stdio.h>
#include <string.h>

// A utility function to get max of two integers
int max(int x, int y) { return (x > y) ? x : y; }

// Returns the length of the longest palindromic subsequence in seq
int lps(char* seq, int i, int j)
{
    // Base Case 1: If there is only 1 character
    if (i == j)
        return 1;

    // Base Case 2: If there are only 2 characters and both are same
    if (seq[i] == seq[j] && i + 1 == j)
        return 2;

    // If the first and last characters match
    if (seq[i] == seq[j])
        return lps(seq, i + 1, j - 1) + 2;

    // If the first and last characters do not match
    return max(lps(seq, i, j - 1), lps(seq, i + 1, j));
}

/* Driver program to test above functions */
int main()
{
    char seq[] = "GEEKSFORGEEKS";
    int n = strlen(seq);
    printf("The length of the LPS is %d", lps(seq, 0, n - 1));
    getchar();
    return 0;
}

Output:

The length of the LPS is 5

Dynamic Programming Solution

C

// A Dynamic Programming based C program for LPS problem
// Returns the length of the longest palindromic subsequence in seq
#include <stdio.h>
#include <string.h>

// A utility function to get max of two integers
int max(int x, int y) { return (x > y) ? x : y; }

// Returns the length of the longest palindromic subsequence in seq
int lps(char* str)
{
    int n = strlen(str);
    int i, j, cl;
    int L[n][n]; // Create a table to store results of subproblems

    // Strings of length 1 are palindrome of length 1
    for (i = 0; i < n; i++)
        L[i][i] = 1;

    // Build the table. Note that the lower diagonal values of table are
    // useless and not filled in the process. The values are filled in a
    // manner similar to Matrix Chain Multiplication DP solution (See
    // https:// www.geeksforgeeks.org/matrix-chain-multiplication-dp-8/). cl is length of
    // substring
    for (cl = 2; cl <= n; cl++) {
        for (i = 0; i < n - cl + 1; i++) {
            j = i + cl - 1;
            if (str[i] == str[j] && cl == 2)
                L[i][j] = 2;
            else if (str[i] == str[j])
                L[i][j] = L[i + 1][j - 1] + 2;
            else
                L[i][j] = max(L[i][j - 1], L[i + 1][j]);
        }
    }

    return L[0][n - 1];
}

/* Driver program to test above functions */
int main()
{
    char seq[] = "GEEKS FOR GEEKS";
    int n = strlen(seq);
    printf("The length of the LPS is %d", lps(seq));
    getchar();
    return 0;
}

Output

The length of the LPS is 7

Time Complexity: O(n²)

Auxiliary Space: O(n²)

Please refer complete article on Longest Palindromic Subsequence | DP-12 for more details!